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当 A 的子数组 A[i], A[i+1], …, A[j] 满足下列条件时，我们称其为湍流子数组：若 i &amp;lt;= k &amp;lt; j，当 k 为奇数时,"> 
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        <h1 class="title">leetcode978.最长湍流子数组</h1>
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            <span>二月 21, 2020</span>
            

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            <h1 id="leetcode978-最长湍流子数组"><a href="#leetcode978-最长湍流子数组" class="headerlink" title="leetcode978. 最长湍流子数组"></a>leetcode978. 最长湍流子数组</h1><hr>
<blockquote>
<p>当 A 的子数组 A[i], A[i+1], …, A[j] 满足下列条件时，我们称其为湍流子数组：<br>若 i &lt;= k &lt; j，当 k 为奇数时， A[k] &gt; A[k+1]，且当 k 为偶数时，A[k] &lt; A[k+1]；<br>或 若 i &lt;= k &lt; j，当 k 为偶数时，A[k] &gt; A[k+1] ，且当 k 为奇数时， A[k] &lt; A[k+1]。<br>也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。<br>返回 A 的最大湍流子数组的长度。</p>
</blockquote>
<blockquote>
<p>示例 1：<br>输入：[9,4,2,10,7,8,8,1,9]<br>输出：5<br>解释：(A[1] &gt; A[2] &lt; A[3] &gt; A[4] &lt; A[5])</p>
</blockquote>
<blockquote>
<p>示例 2：<br>输入：[4,8,12,16]<br>输出：2</p>
</blockquote>
<blockquote>
<p>示例 3：<br>输入：[100]<br>输出：1</p>
</blockquote>
<p>我是用滑动窗口解的，因为最近在练这个。右边窗口移动的条件：前一个是小于，这次比较是大于或者前一个是大于，这次比较是小于，用flag记录前一次比较是什么情况；还有一种情况是前一次不满足湍流数组，使flag等于0.经过上述判断后还不满足的，就表示当前窗口内部不是一个湍流数组，需要移动i，把i移动到j的前一个，对于 <strong>[9,4,2,10,7,8,8,1,9]</strong> ,第一次修改i，i会到4的位置，并且下一次的j需要从2开始。但是这么写会在之后的 <strong>8,8</strong>往复循环，所以在还要判断一个是否相等的情况。没有任何优化，看着也很丑，不过思路还算比较清晰的。10ms.</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">maxTurbulenceSize</span><span class="params">(<span class="keyword">int</span>[] A)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(A.length==<span class="number">1</span>)<span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">int</span> flag=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> result=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>,j=<span class="number">1</span>;j&lt;A.length;j++) &#123;</span><br><span class="line">        	<span class="keyword">if</span>((A[j]&gt;A[j-<span class="number">1</span>]&amp;&amp;flag==<span class="number">1</span>)||(A[j]&gt;A[j-<span class="number">1</span>]&amp;&amp;flag==<span class="number">0</span>))</span><br><span class="line">        		flag=-<span class="number">1</span>;</span><br><span class="line">        	<span class="keyword">else</span> <span class="keyword">if</span>((A[j]&lt;A[j-<span class="number">1</span>]&amp;&amp;flag==-<span class="number">1</span>)||(A[j]&lt;A[j-<span class="number">1</span>]&amp;&amp;flag==<span class="number">0</span>))</span><br><span class="line">        		flag=<span class="number">1</span>;</span><br><span class="line">        	<span class="keyword">else</span> &#123;</span><br><span class="line">        		<span class="keyword">if</span>(A[j]==A[j-<span class="number">1</span>]) &#123;</span><br><span class="line">        			i=j;</span><br><span class="line">                	result=Math.max(j-i+<span class="number">1</span>, result);</span><br><span class="line">        			<span class="keyword">continue</span>;</span><br><span class="line">        		&#125;</span><br><span class="line">        		i=j-<span class="number">1</span>;</span><br><span class="line">        		j--;</span><br><span class="line">        		flag=<span class="number">0</span>;</span><br><span class="line">        	&#125;</span><br><span class="line">        	result=Math.max(j-i+<span class="number">1</span>, result);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> result;</span><br><span class="line">    &#125;</span><br></pre></td></tr></table></figure>
<p><strong>leetcode 7/100</strong></p>

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